14. Centralizers (continued)

Our goal is to understand the abelian subgroups of a hyperbolic group; e.g., can a hyperbolic group contain as a subgroup? We already know that it cannot be a quasiconvex subgroup, but it may be possible for to be “twisted” in some manner.

Theorem 10. The intersection of two quasiconvex subgroups is quasiconvex.

Pf. As usual, we work in a -hyperbolic . Let be quasiconvex subgroups each with the same corresponding constant . Let , and let . Our goal is therefore to show that is in a bounded neighborhood of . Let be the (or more precisely, a particular) closest element of to , and suppose that . Let be such that and let be such that ; such elements exist since and are quasiconvex. Let be such that . We sketch the situation below.

Consider the geodesic triangle with vertices , , and . Because this triangle is -slim, for each there exists some such that .

Likewise, for each there exists such that . Next, let and . Then .

Suppose . Then by the Pigeonhole Principle there are integers and with such that and . Now, we can use this information to find a closer element of .

Consider . By the triangle inequality,

.

All that remains is to prove that . But since ,

.

Playing this same game with , we get , and hence we have found our contradiction.

For a group , recall that is the center of .

Corollary. For any (where is -hyperbolic) of infinite order, the subgroup generated by is quasiconvex. Equivalently, the map sending is a quasigeodesic (which is a sensible statement to make given that is quasi-isometric to ).

Pf. By Theorem 9, we deduce that is quasiconvex, and so is finitely generated. Let be a finite generating set for . Notice

where is the centralizer in of , so is quasiconvex by Theorem 10 and thus is a finitely generated abelian group containing . By Exercise 16 (below), we deduce that is quasi-isometrically embedded in , and hence is quasiconvex in .